Sketch the parametrised curve $C = \{(1-e^{-t},e^{-2t}):t\in [0,\infty]\}$

Question

Exercise: Sketch the parametrised curve $C = \{(1-e^{-t},e^{-2t}):t\in [0,\infty]\}$.

What I've tried: If you let $f(t) = \begin{bmatrix}1 - e^{-t}\\e^{-2t}\end{bmatrix}$ then for $t = 0,1,2,\ldots$ you get $$f(0) = \begin{bmatrix}0\\1\end{bmatrix}, \,f(1) = \begin{bmatrix}1 - e^{-1}\\e^{-2}\end{bmatrix},\,f(2) = \begin{bmatrix}1 - e^{-2}\\e^{-4}\end{bmatrix},\ldots$$

Since I wasn't able to make anything of this myself, I used the Wolfram parametric curve plotter, but it said that I didn't enter a valid input.

Question: How do I sketch this parametrised curve? Why is wolfram telling me that this input is incorrect? Surely I'm doing something wrong, but I can't figure out what it is.

Thanks!

Answer 1

Hint:

$$x(t)= 1-\exp(-t)$$ $$y(t)=\exp(-2t)=\left[\exp(-t) \right]^2$$

Solve the first equation for $\exp(-t)$ and plug it into the second equation to obtain $y(x)$.

Also, think about all possible values of $x$? What domain is admissible for $x$?